Review Conceptual Example 6 before working this problem. For water vapor in air

Question

Review Conceptual Example 6 before working this problem. For water vapor in air a 293 K, the diffusion constant is D = 2.4 Times 10^-5 m^2/s. As outlined in Problem 51(a), the time required for the first solute molecules to traverse a channel of length L is t = L^2/(2D), according to Fick's law. Find the time t for L = 0.0216 m. For comparison, how long would a water molecule take to travel L = 0.0216 m at the translational rms speed of water molecules (assumed to be an ideal gas) at a temperature of 293 K? Number Units Number Units

Explanation / Answer

Diffusion constant D=2.4×10^-5 m^2/s

(A) ans

Length L=0.0216 m

We know that the time required t=L^2/2D

=(0.0216)^2/2×2.4×10^-5

=4.7×10^-4/4.8×10^-5

=9.8 sec

(B) ans

Now we find the RMS velocity Vrms=[3RT/M]

RMS velocity =[3×8.1×293/36]

=[7119.9/36]

=14.1 m/s

therefore the time t=s/v=0.0216/14.1=1.532*10^-3 sec

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